In the following, find the digits represented by the letters:

(i) (ii)

(iii) (iv)

(v) (vi)

(i) Here B = 4 as when we subtract 3 from 7 that is 7-3 = 4 and 3 + 4 = 7

(ii) here A = 5 as only 6 + 5 = 11

From 11 the one`s place 1 has been written down, and another ten`s place 1 has been carried over to the ten`s place of given question

So the value of B = 4

(iii) Here A is occurring at all the three places (in the question and answer)

⇒ A could take values1, 5 and 6

If we take A = 1 then the answer would be 21 × 1 = 21 which is not in accordance the given solution

If A = 6 then the sum becomes 26 × 6 = 156 which is not the solution

So A = 5

As 25 × 5 = 125

_{(iv) Here all the A are the same in the question as well as answer}

_{So A can take the only value as 0 as 0 + 0 = 0}

_{Which is the only solution?}

(v) here A is coming in question and answer both so

Values A can have are 0 and 1

With A = 0 the sum becomes 10 × 10 = 100

So if the value of A = 0 then B = 0

And if we take value of A = 1, the sum is 11 × 11 = 121

So if A = 1 and then B = 2

_{(iv) Here A can take value of 1, 5 or 6}

Now if A = 1 then the multiplication would be 31 × 1 = 31 which is not the solution

If A = 5 the multiplication would be 35 × 5 = 175 which is not the solution again

If A = 6 then the multiplication becomes 36 × 6 = 216

which satisfies the condition of the solution

_{⇒ A = 6 and B = 1}